\(\int e^{-3 i \arctan (a+b x)} \, dx\) [211]

Optimal result

Integrand size = 12, antiderivative size = 94 \[ \int e^{-3 i \arctan (a+b x)} \, dx=\frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-\frac {3 \text {arcsinh}(a+b x)}{b} \]

[Out]

-3*arcsinh(b*x+a)/b+2*I*(1-I*a-I*b*x)^(3/2)/b/(1+I*a+I*b*x)^(1/2)+3*I*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/
b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5201, 49, 52, 55, 633, 221} \[ \int e^{-3 i \arctan (a+b x)} \, dx=-\frac {3 \text {arcsinh}(a+b x)}{b}+\frac {2 i (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}+\frac {3 i \sqrt {i a+i b x+1} \sqrt {-i a-i b x+1}}{b} \]

[In]

Int[E^((-3*I)*ArcTan[a + b*x]),x]

[Out]

((2*I)*(1 - I*a - I*b*x)^(3/2))/(b*Sqrt[1 + I*a + I*b*x]) + ((3*I)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x]
)/b - (3*ArcSinh[a + b*x])/b

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 55

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 5201

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.)), x_Symbol] :> Int[(1 - I*a*c - I*b*c*x)^(I*(n/2))/(1 + I*a*c +
I*b*c*x)^(I*(n/2)), x] /; FreeQ[{a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(1-i a-i b x)^{3/2}}{(1+i a+i b x)^{3/2}} \, dx \\ & = \frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}-3 \int \frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx \\ & = \frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-3 \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx \\ & = \frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-3 \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx \\ & = \frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )}{2 b^2} \\ & = \frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-\frac {3 \text {arcsinh}(a+b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.48 \[ \int e^{-3 i \arctan (a+b x)} \, dx=\frac {\sqrt {1+(a+b x)^2} \left (i+\frac {4}{-i+a+b x}\right )}{b}-\frac {3 \text {arcsinh}(a+b x)}{b} \]

[In]

Integrate[E^((-3*I)*ArcTan[a + b*x]),x]

[Out]

(Sqrt[1 + (a + b*x)^2]*(I + 4/(-I + a + b*x)))/b - (3*ArcSinh[a + b*x])/b

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.37

method	result	size
risch	\(\frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b}-\frac {3 \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}+\frac {4 \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{b^{2} \left (x -\frac {i-a}{b}\right )}\)	\(129\)
default	\(\frac {i \left (\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{3}}-2 i b \left (-\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{2}}+3 i b \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{2 \sqrt {b^{2}}}\right )\right )\right )\right )}{b^{3}}\)	\(327\)

[In]

int(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

I/b*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)+4/b^
2/(x-(I-a)/b)*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int e^{-3 i \arctan (a+b x)} \, dx=\frac {{\left (i \, a + 8\right )} b x + i \, a^{2} + 6 \, {\left (b x + a - i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-i \, b x - i \, a - 5\right )} + 9 \, a - 8 i}{2 \, {\left (b^{2} x + {\left (a - i\right )} b\right )}} \]

[In]

integrate(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*((I*a + 8)*b*x + I*a^2 + 6*(b*x + a - I)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - 2*sqrt(b^2*x^
2 + 2*a*b*x + a^2 + 1)*(-I*b*x - I*a - 5) + 9*a - 8*I)/(b^2*x + (a - I)*b)

Sympy [F]

\[ \int e^{-3 i \arctan (a+b x)} \, dx=i \left (\int \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx + \int \frac {a^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx + \int \frac {b^{2} x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx + \int \frac {2 a b x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx\right ) \]

[In]

integrate(1/(1+I*(b*x+a))**3*(1+(b*x+a)**2)**(3/2),x)

[Out]

I*(Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 + 3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3
*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x) + Integral(a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 + 3
*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x) + Integral
(b**2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 + 3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*
a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x) + Integral(2*a*b*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 +
3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int e^{-3 i \arctan (a+b x)} \, dx=\frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b - 2 i \, b^{2} x - 2 i \, a b - b} - \frac {3 \, \operatorname {arsinh}\left (b x + a\right )}{b} + \frac {6 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{i \, b^{2} x + i \, a b + b} \]

[In]

integrate(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="maxima")

[Out]

I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/(b^3*x^2 + 2*a*b^2*x + a^2*b - 2*I*b^2*x - 2*I*a*b - b) - 3*arcsinh(b*x
+ a)/b + 6*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(I*b^2*x + I*a*b + b)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (66) = 132\).

Time = 0.31 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.91 \[ \int e^{-3 i \arctan (a+b x)} \, dx=\frac {\log \left (3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b + a^{3} b + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} {\left | b \right |} + 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} {\left | b \right |} - 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b - 2 i \, a^{2} b + 4 \, {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} - a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{{\left | b \right |}} + \frac {i \, \sqrt {{\left (b x + a\right )}^{2} + 1}}{b} \]

[In]

integrate(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="giac")

[Out]

log(3*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a*b + a^3*b + (x*abs(b) - sqrt((b*x + a)^2 + 1))^3*abs(b) + 3*(x*ab
s(b) - sqrt((b*x + a)^2 + 1))*a^2*abs(b) - 2*I*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*b - 2*I*a^2*b + 4*(-I*x*ab
s(b) + I*sqrt((b*x + a)^2 + 1))*a*abs(b) - a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/abs(b) + I*sqrt((b
*x + a)^2 + 1)/b

Mupad [F(-1)]

Timed out. \[ \int e^{-3 i \arctan (a+b x)} \, dx=\int \frac {{\left ({\left (a+b\,x\right )}^2+1\right )}^{3/2}}{{\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

[In]

int(((a + b*x)^2 + 1)^(3/2)/(a*1i + b*x*1i + 1)^3,x)

[Out]

int(((a + b*x)^2 + 1)^(3/2)/(a*1i + b*x*1i + 1)^3, x)